The Cauchy product of ${\sum }_{k\ge 0}(-1{)}^k\mathrm{/}\sqrt{k+1}$ with itself has $\mathrm{\mid }{c}_n\mathrm{\mid }\ge 1$ for every $n$, so it diverges

Refuted claim: the Cauchy product of two convergent series of reals converges (The Cauchy product of two series: $c_n={\sum }_{k=0}^n{a}_k{b}_{n-k}$, Series, partial sums, convergence and the sum, divergence, and the tail series).

The witness is a single conditionally convergent series multiplied by itself. Let $({\epsilon }_k)$ be the alternating sequence (The even and odd index maps and the alternating sequence: strictly increasing $e,o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0=1$, $s_{\sigma (k)}=-s_k$, which satisfies $\mathrm{\mid }{s}_k\mathrm{\mid }=1$, $s\circ e\equiv 1$ and $s\circ o\equiv -1$) and

$a_k=b_k:=\frac{{\epsilon }_k}{\sqrt{\iota (k+1)}},$

so that $\sum a_k$ converges by the alternating series test. Then, as FALSE: the Cauchy product of two convergent series converges establishes,

$\mathrm{\mid }{c}_n\mathrm{\mid }={\sum }_{k=0}^n\frac{1}{\sqrt{\iota (k+1)\iota (n-k+1)}}\ge \frac{2\iota (n+1)}{\iota (n+2)}\ge 1(n\in \mathbb{N}),$

so $(c_n)$ does not converge to $0$ and $\sum c_n$ diverges (If a series converges then its terms tend to $0$).

What this counterexample adds to the false statement is the sharp form of the bound: the lower bound $2\iota (n+1)\mathrm{/}\iota (n+2)=2-2\mathrm{/}\iota (n+2)$ increases to $2$, so $\mathrm{\mid }{c}_n\mathrm{\mid }$ eventually exceeds every real below $2$. The terms of the product series therefore do not merely fail to tend to $0$; they stay bounded away from it by an amount approaching $2$. Nothing here determines the asymptotic size of $\mathrm{\mid }{c}_n\mathrm{\mid }$ itself, only this lower bound for it; the divergence is as far from marginal as the bound makes it.

• Why the product cannot cancel. By The even and odd index maps and the alternating sequence: strictly increasing $e,o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0=1$, $s_{\sigma (k)}=-s_k$, which satisfies $\mathrm{\mid }{s}_k\mathrm{\mid }=1$, $s\circ e\equiv 1$ and $s\circ o\equiv -1$ the sign of $a_k{b}_{n-k}$ is ${\epsilon }_k{\epsilon }_{n-k}={\epsilon }_n$, the same for every $k\le n$: the antidiagonal of the product array is sign-constant. So all $n+1$ terms of $c_n$ add, and the AM-GM bound (The arithmetic mean, geometric mean inequality, with square roots as in Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$) shows each is at least $2\mathrm{/}\iota (n+2)$.

• Compare the geometric case. In For $\mathrm{\mid }r\mathrm{\mid }<1$ the Cauchy product of $\sum r^k$ with itself is $\sum (k+1)r^k$, with sum $1\mathrm{/}(1-r{)}^2$ the antidiagonal also has $n+1$ equal terms, but they are $r^n$ with $\mathrm{\mid }r\mathrm{\mid }<1$, so the count $n+1$ is beaten by the decay. Here the terms of the antidiagonal are at least $2\mathrm{/}\iota (n+2)$ each, and there are $n+1$ of them, so the count wins.

• This is the same series as in With $a_j=(-1{)}^j\mathrm{/}\sqrt{j+1}$ convergent and $b_j=(-1{)}^j$ bounded but not monotone, $\sum a_j{b}_j=\sum 1\mathrm{/}\sqrt{j+1}$ diverges. Its terms are exactly small enough to be summable with alternating signs and not otherwise, which is what makes it the standard witness for both failures.